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Math Help - Solution of equation of form ln(a+bx) = cx

  1. #1
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    Solution of equation of form ln(a+bx) = cx

    Hello,

    I am having an exponential equation that depends on 3 different variables, which are included in the constants a, b and c....
    solving this equation, I get something of the form of ln(a + bx) - cx = 0

    I don't manage to simplify the equation any further
    I was wondering whether this is actually possible without any numerical method, I would like to get an equation for x as a function of a, b and c (or actually some other constants that form a, b and c).

    Or is there maybe any approximations method , such as a Taylor development or something, which gives me a solution with a certain accuracy?

    Would be nice if someone could help me!

    Cheers
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  2. #2
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    Re: Solution of equation of form ln(a+bx) = cx

    i think you need to use the Lambert W-function here: Lambert W-Function -- from Wolfram MathWorld
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  3. #3
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    Re: Solution of equation of form ln(a+bx) = cx

    Thank you very much for your answer!

    I had already heard about the Lambert W-function in order to solve problems like this, but I have to admit that I don't really understand how that works... (I'm not tha good in maths, I'm just a chemist )
    I understood what the Lambert function is but I don't manage to apply it to solve my problem :S

    Could you help me, please?

    Thanks a lot
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  4. #4
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    Re: Solution of equation of form ln(a+bx) = cx

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  5. #5
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    Re: Solution of equation of form ln(a+bx) = cx

    \displaystyle a+bx=e^{cx}

    \displaystyle (a+bx)e^{-cx}=1

    \displaystyle -(\frac{ac}{b}+cx)e^{-cx}=-\frac{c}{b}

    \displaystyle -(\frac{ac}{b}+cx)e^{-(ac/b+cx)}=-\frac{c}{b}e^{-ac/b}


    This is now ready to apply the Lambert function...

    Pretty sure I've got a slip in there somewhere so look out.
    Last edited by a tutor; April 18th 2012 at 05:09 AM.
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  6. #6
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    Re: Solution of equation of form ln(a+bx) = cx

    Thanks a lot
    But I still don't know how to APPLY the Lambert-W function... and so how to get a solution for x....
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  7. #7
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    Re: Solution of equation of form ln(a+bx) = cx

    Dude forget that Lambert W(hatever) stuff.

    Here goes:

    First plot the functions g(x)=ln(a+bx) and h(x)=cx, both in the same coordinate plane. Abscissas of the intersection points are solutions to the equation g(x)=h(x) or ln(a+bx)=cx. Now from that picture you can estimate the intervals that contain the solutions. Say that interval is I=[u,v].

    Now you just have state the precision you need (epsilon=0.00005 would mean 4 significant decimal digits) and use an iterative method that would find the solution in the interval I=[u,v]. Good choice of method would be Newton tangent method, or even plain bisection method provided you don't need great precision.
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  8. #8
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    Re: Solution of equation of form ln(a+bx) = cx

    \displaystyle W\left(-\left(\frac{ac}{b}+cx\right)e^{-(ac/b+cx)}\right)=W\left(-\frac{c}{b}e^{-ac/b}\right)

    \displaystyle -\left(\frac{ac}{b}+cx\right)=W\left(-\frac{c}{b}e^{-ac/b}\right)

    \displaystyle \frac{ac}{b}+cx\right=-W\left(-\frac{c}{b}e^{-ac/b}\right)

    \displaystyle cx=-W\left(-\frac{c}{b}e^{-ac/b}\right)-\frac{ac}{b}

    \displaystyle x=-\frac{1}{c}W\left(-\frac{c}{b}e^{-ac/b}\right)-\frac{a}{b}
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